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 April 12th, 2012, 11:01 PM #1 Member   Joined: Apr 2012 Posts: 92 Thanks: 0 Inequalities, number theory Hello all, I have a question and it may seem a little weird that I should ask it, but I really am confused. You see, I have just completed a topic on inequalities and was looking at number theory and found that the things I had learned in inequalities did not apply here, or so it seems. Find the sum of n terms of the GP 3+4.5+6.75+... How large must n be so that this sum is greater than 6000? I was able to identify common ratio r to be 1.5 Therefore, Sn>6000 Sn = 3(1-(1.5)^n) / 1-1.5 >6000 3(1-(1.5)^n) / 0.5 > 6000 -6(1-(1.5)^n)>6000 Because I cannot being over a negative with an inequality present, 6((1.5)^n -1) > 6000 ((1.5)^n -1) > 1000 And so on till I got n> log1.5(1001) n> 17.03, Since I cannot round down, n>18 ( Answer ) Was my working right? The answer was n>=18, can anyone tell me why? 6((1.5)^n -1) > 6000 ((1.5)^n -1) > 1000 This is the part where I'm confused. I vaguely remember that this is how you should approach this problem but what I learned in inequalities told me to in this case, square both sides since LHS and RHS are positive. OHHH actually I just remembered. I can bring over 6 because it is definitely positive right?
 April 12th, 2012, 11:24 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Inequalities, number theory I would write: $\sum_{k=0}^{n-1}3\cdot$$\frac{3}{2}$$^k=\frac{3$$1-\(\frac{3}{2}$$^{n}\)}{1-\frac{3}{2}}=6$$\(\frac{3}{2}$$^{n}-1\)$ $6$$\(\frac{3}{2}$$^{n}-1\)>6000$ $$$\frac{3}{2}$$^{n}-1>1000$ $$$\frac{3}{2}$$^{n}>1001$ $\ln$$\(\frac{3}{2}$$^{n}\)>\ln$$1001$$$ $n\ln$$\frac{3}{2}$$>\ln$$1001$$$ $n>\frac{\ln(1001)}{\ln$$\frac{3}{2}$$}$ $n>\frac{\ln(1001)}{\ln$$\frac{3}{2}$$}$ $n>17.03908583293485$ $n\ge18$ You use a weak inequality because n can be 18 or greater. Your working was correct, you just needed a weak inequality at the end.
 April 13th, 2012, 01:23 AM #3 Member   Joined: Apr 2012 Posts: 92 Thanks: 0 Re: Inequalities, number theory How did you get n-1 and what is a weak inequality? Thank you!
 April 13th, 2012, 05:06 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Inequalities, number theory I let the summation go from k = 0 to k = n - 1, which is n terms. A strict inequality is strictly less than or greater than, while a weak inequality is less than or equal to or greater than or equal to.

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